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UPSC CDS I 2019 Math Paper

Question Numbers: 94-96

Consider the following for the next three (03) items :

ABCD is a quadrilateral with A

B = 9

cm, B

C = 40

cm, C

D = 28

cm, D

A = 15

cm and angle ABC is a right-angle.

Question : 95 of 100

Marks: +1, -0

What is the area of quadrilateral ABCD?

Solution:

Concept:

The area of quadrilateral

ABCD

is found by splitting it into two triangles and summing their areas.

Explanation:

In triangle

ABC

\angle ABC = 90^{\circ}

AB = 9

cm,

BC = 40

cm.

By Pythagoras theorem:

AC^2 = AB^2 + BC^2 = 9^2 + 40^2 = 81 + 1600 = 1681

AC = \sqrt{1681} = 41

cm.

Area of

\triangle ABC = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 40 \times 9 = 180

^{2}

For triangle

ACD

, sides are

AC = 41

cm,

CD = 28

cm,

DA = 15

cm.

Semi-perimeter

s = \frac{41 + 28 + 15}{2} = 42

cm.

Using Heron's formula:

Area

= \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{42 \times (42-41) \times (42-28) \times (42-15)}

= \sqrt{42 \times 1 \times 14 \times 27} = \sqrt{15876} = 126

^{2}

Area of quadrilateral

ABCD

= area of

\triangle ABC

+ area of

\triangle ACD

180 + 126 = 306

^{2}

Answer:

306

^{2}

(Option B).

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